16(3x-3)=-18x(x+10)

Solution for 16(3x-3)=-18x(x+10) equation:

16(3x-3)=-18x(x+10)

We move all terms to the left:

16(3x-3)-(-18x(x+10))=0

We multiply parentheses

48x-(-18x(x+10))-48=0


We calculate terms in parentheses: -(-18x(x+10)), so:

-18x(x+10)

We multiply parentheses

-18x^2-180x

Back to the equation:

-(-18x^2-180x)


We get rid of parentheses

18x^2+180x+48x-48=0

We add all the numbers together, and all the variables

18x^2+228x-48=0

a = 18; b = 228; c = -48;
Δ = b2-4ac
Δ = 2282-4·18·(-48)
Δ = 55440
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{55440}=\sqrt{144*385}=\sqrt{144}*\sqrt{385}=12\sqrt{385}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(228)-12\sqrt{385}}{2*18}=\frac{-228-12\sqrt{385}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(228)+12\sqrt{385}}{2*18}=\frac{-228+12\sqrt{385}}{36}
$